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The given problem asks to determine whether the series

\[
\sum_{k=2}^{\infty} \sqrt{\frac{k^2 - k}{k^4}}
\]

converges or diverges using the **Limit Comparison Test**.

### Step-by-step Solution:

1. **Simplify the given term**:

   First, simplify the expression inside the summation:

   \[
   \sqrt{\frac{k^2 - k}{k^4}} = \sqrt{\frac{k^2(1 - \frac{1}{k})}{k^4}} = \sqrt{\frac{1 - \frac{1}{k}}{k^2}} = \frac{\sqrt{1 - \frac{1}{k}}}{k}
   \]

   As \( k \) becomes large, \(\frac{1}{k}\) becomes small, so the expression behaves asymptotically like:

   \[
   \frac{1}{k^{3/2}}.
   \]

2. **Comparison Series**:

   Based on the asymptotic behavior, a good candidate for the comparison series is:

   \[
   \sum_{k=2}^{\infty} \frac{1}{k^{3/2}}.
   \]

   This is a **p-series** with \( p = \frac{3}{2} \). Since \( p > 1 \), the p-series converges.

3. **Limit Comparison Test**:

   To apply the Limit Comparison Test, we compute the limit of the ratio of the terms of the given series and the comparison series:

   \[
   \lim_{k \to \infty} \frac{\frac{\sqrt{1 - \frac{1}{k}}}{k}}{\frac{1}{k^{3/2}}} = \lim_{k \to \infty} \frac{k^{3/2}}{k} \cdot \sqrt{1 - \frac{1}{k}} = \lim_{k \to \infty} k^{1/2} \cdot \sqrt{1 - \frac{1}{k}}.
   \]

   As \( k \to \infty \), \(\sqrt{1 - \frac{1}{k}} \to 1\), so:

   \[
   \lim_{k \to \infty} k^{1/2} \cdot 1 = 1.
   \]

   Since this limit is a finite, non-zero constant (specifically, \(1\)), the Limit Comparison Test tells us that the given series behaves like the comparison series, which converges.

### Conclusion:

Therefore, by the Limit Comparison Test, the series

\[
\sum_{k=2}^{\infty} \sqrt{\frac{k^2 - k}{k^4}}
\]

**converges**.

---

Do you want further details or have any questions on this solution? 

Here are 5 related questions to deepen your understanding:

1. What are the conditions for applying the Limit Comparison Test?
2. How would the conclusion change if the comparison series diverged instead?
3. Can you apply the Direct Comparison Test here, and would it lead to the same result?
4. What happens if we try to approximate the given series for large \( k \) before applying the test?
5. What are other types of convergence tests you could use for this series?

**Tip:** The Limit Comparison Test works best when the given series has similar asymptotic behavior to a known p-series or geometric series.

Use the Limit Comparison Test to determine whether the series \( \sum_{k=2}^{\infty} \sqrt{\frac{k^2 - k}{k^4}} \) converges or diverges.

Determine whether the series \( \sum_{k=2}^{\infty} \sqrt{\frac{k^2 - k}{k^4}} \) converges or diverges using the Limit Comparison Test. This problem covers series convergence, asymptotic behavior, and the application of the Limit Comparison Test, suitable for college-level calculus students.

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